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Let $f\left( z \right) = \sin z$ and $g\left( z \right) = \cos z$ . If $*$ denotes a composition of the function then show that $\left( {f + ig} \right) * \left( {f - ig} \right) = i{e^{ - {e^{iz}}}}$ . $i = \sqrt { - 1}$

Last updated date: 13th Jun 2024
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Hint: Since we have to show $\left( {f + ig} \right) * \left( {f - ig} \right) = i{e^{ - {e^{iz}}}}$ and for this we will write $\left( {f + ig} \right) * \left( {f - ig} \right)$ in the composition form and it will be $F\left( {G\left( z \right)} \right)$ . From here $F\left( z \right) = f + ig$ and $G\left( z \right) = f - ig$ . Then we will solve both the composite equation and then in the end putting the values we will get the solution.
Formula used:
The algebraic formula is given by,
$\cos z - i\sin z = {e^{ - iz}}$
$i = \sqrt { - 1}$
Here, $i$ will be the iota.

Here, in this question, we have the values of the function given as $f\left( z \right) = \sin z$ and $g\left( z \right) = \cos z$ .
And we have to prove $\left( {f + ig} \right) * \left( {f - ig} \right) = i{e^{ - {e^{iz}}}}$
Now taking the LHS of the upper function, we get
$\Rightarrow \left( {f + ig} \right) * \left( {f - ig} \right)$
And it can also be written in the composition form, so it will be
$\Rightarrow F\left( z \right) * G\left( z \right)$
And it will be equal to
$\Rightarrow F\left( {G\left( z \right)} \right)$
Since, $F\left( z \right) = f + ig$
So, on substituting the values, we get
$\Rightarrow F\left( z \right) = \left( {\sin z + i\cos z} \right)$
Multiplying and dividing the above equation with the same function, $i$
We get,
$\Rightarrow F\left( z \right) = \left( {\sin z + i\cos z} \right)\dfrac{i}{i}$
And on solving it, we get
$\Rightarrow F\left( z \right) = \dfrac{{i\sin z + {i^2}\cos z}}{i}$
Since we know $i = \sqrt { - 1}$ , hence
$\Rightarrow F\left( z \right) = \dfrac{{i\sin z - \cos z}}{i}$
Taking the negative sign common, we get
$\Rightarrow F\left( z \right) = - \dfrac{1}{i}\left( {\cos z - i\sin z} \right)$
Also, from the formula, we know that $\cos z - i\sin z = {e^{ - iz}}$ . So by using it we get
$\Rightarrow F\left( z \right) = - \dfrac{1}{i}{e^{ - iz}}$
Again multiplying and dividing the above equation with the same function, $i$
We get
$\Rightarrow F\left( z \right) = - \dfrac{1}{i} \cdot \dfrac{i}{i}{e^{ - iz}}$
And on solving it we get
$\Rightarrow F\left( z \right) = - \dfrac{{1 \cdot i}}{{{i^2}}}{e^{ - iz}}$
Hence, it will be equal to
$\Rightarrow F\left( z \right) = i{e^{ - iz}}$
Now we will solve, $G\left( z \right) = f - ig$
So, on substituting the values, we get
$\Rightarrow G\left( z \right) = \left( {\sin z - i\cos z} \right)$
Multiplying and dividing the above equation with the same function, $i$
We get,
$\Rightarrow G\left( z \right) = \left( {\sin z - i\cos z} \right)\dfrac{i}{i}$
And on solving it, we get
$\Rightarrow G\left( z \right) = \dfrac{{i\sin z - {i^2}\cos z}}{i}$
Since we know $i = \sqrt { - 1}$ , hence
$\Rightarrow G\left( z \right) = \dfrac{{i\sin z + \cos z}}{i}$
Also, from the formula, we know that $\cos z - i\sin z = {e^{ - iz}}$ . So by using it we get
$\Rightarrow G\left( z \right) = \dfrac{{{e^{iz}}}}{i}$
Again multiplying and dividing the above equation with the same function, $i$
We get
$\Rightarrow G\left( z \right) = - \dfrac{1}{i} \cdot \dfrac{i}{i}{e^{iz}}$
And on solving it we get
$\Rightarrow G\left( z \right) = - \dfrac{{1 \cdot i}}{{ - 1}}{e^{iz}}$
Hence, it will be equal to
$\Rightarrow G\left( z \right) = - i{e^{iz}}$
So from the upper part of the solution,
$\Rightarrow F\left( {G\left( z \right)} \right) = i{e^{i \cdot \left( {G\left( z \right)} \right)}}$
And on substituting the values, we get
$\Rightarrow F\left( {G\left( z \right)} \right) = i{e^{i \cdot \left( { - i{e^{iz}}} \right)}}$
By using identities, we get the equation as
$\Rightarrow F\left( {G\left( z \right)} \right) = i{e^{ - {i^2}{e^{iz}}}}$
And it can also be written as
$\Rightarrow F\left( {G\left( z \right)} \right) = i{e^{{e^{iz}}}}$
Since, we can see that LHS is equal to RHS,
Therefore, it is proved that $\left( {f + ig} \right) * \left( {f - ig} \right) = i{e^{ - {e^{iz}}}}$ will be equal.
So, the correct answer is “ $\left( {f + ig} \right) * \left( {f - ig} \right) = i{e^{ - {e^{iz}}}}$ ”.

Note: Solving composite functions means we have to find the composition of two functions. By rewriting the composite function in a different form we will be able to reach close to the answer. And lastly, by substituting the values we get it.